luogu1034[矩形覆盖]

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dp[i][k]表示i个点被j个矩形覆盖的最小面积
s[i][j]表示第i个点到第j个点的面积
dp[i][k]=dp[j][k-1]+s[j+1][i]

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 50 + 5 ;int s [N] [N] ;int dp [N] [5] ;struct P {    int x , y ;}a [N] ;int cmp ( P l , P r ) {    if ( l.x == r.x ) return l.y < r.y ;    return l.x < r.x ;}int main (){    int n , K ;    scanf ( "%d%d" , &n , &K ) ;    for ( int i = 1 ; i <= n ; ++ i )         scanf ( "%d%d" , & a [i].y , & a [i].x ) ;    sort ( a + 1 , a + 1 + n , cmp );    for ( int i = 1 ; i <= n ; ++ i ){        int h = a [i].y , l = a [i].y ;        for ( int j = i ; j <= n ; ++ j ) {            h = max ( h , a [j].y ) ;            l = min ( l , a [j].y ) ;            s [i] [j] = (h-l)*(a[j].x-a[i].x);        }    }    memset ( dp , 127 , sizeof dp ) ;     for ( int i = 1 ; i <= n ; ++ i ){        dp [i] [1] = s [1] [i] ;        for ( int k = 2 ; k <= K ; ++ k )             for ( int j = k-1 ; j < i ; ++ j )                dp [i] [k] = min ( dp [i] [k] , dp [j] [k-1] + s [j+1] [i] );    }     printf ( "%d" , dp [n] [K] ) ;    return 0 ; }

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